Appendix C Doléans measures C.1 Introduction Once again all random processes will live on a fixed probability space (Ω, F, P equipped with a filtration {F t : 0 t 1}. We should probably assume the filtration is standard even though it does not seem essential for this project. The construction (Project 6 of the isometric stochastic integral H M with respect to a martingale M M 2 0 [0, 1], at least for bounded, predictable H, depended on the existence of the Doléans measure µ on the predictable sigma-field P on S 1 := Ω (0, 1]. To make the map H H M 1 an isometry between H simple and a subset of L 2 (Ω, F 1, P we needed <1> µ(a, b] F = P{ω F }(M b M a 2 for all 0 a < b 1 and F F a. This property characterizes the measure µ because the collection of all predictable sets of the form (a, b] F is f-stable and it generates P. The sigma-field P is also generated by the set of all stochastic intervals ((0, τ]] for τ T 1, the set of all [0, 1]-valued stopping times for the filtration. The Doléans measure is also characterized by the property µ((0, τ]] = PM 2 τ for all τ T 1. Notice that µ depends on M only through the submartingale S t := M 2 t. In fact, analogous measures can be defined for a large class of submartingales. <2> Definition. Let {S t : 0 t 1} be a cadlag submartingale with S 0 0. Say that a finite (countably-additive measure µ S, defined on the predictable sigma-field P on S, is the Doléans measure for S if µ((0, τ]] = PS τ for every τ in T 1, the set of all [0, 1]-valued stopping times for the filtration. Remark. If µ S exists then S τ must be integrable for every τ in T 1. In fact we will need the uniform integrability only for stopping times with a finite range (Métivier 1982, page 80. I mentioned explicitly that µ S must be countably-additive to draw attention to a subtle requirement on S for µ S to exist, known somewhat cryptically as property [D]: [D] the set of random variables {S τ : τ T 1 } is uniformly integrable. 27 February 2010 Advanced stochastic processes, spring 2010 c David Pollard
C.2 2 Problem [2] shows that if there exists a (countably-additive Doléans measure µ S then S has property [D]. The proof of the converse assertion is the main subject of this Appendix. <3> Theorem. Every cadlag submartingale {S t : 0 t 1} with property [D] has a countably additive Doléans measure µ S on the predictable sigma-field P on S := Ω (0, 1]. Remark. Alert readers will be thinking correctly that property [D] must be used somewhere to convert a pointwise convergence along a sequence of stopping times into an L 1 convergence. <4> Example. If M M 2 0 [0, 1] then the submartingale S t = M 2 t has property [D]. Indeed, we know from Project 4 that M τ = P(M 1 F τ for all τ T 1. Thus 0 M 2 τ P ( M 2 1 F τ for each τ in T1. Use the fact that {P(ξ G : G a sub-sigma-field of F} is uniformly integrable for each integrable random variable ξ to complete the argument. Remark. If I write µ S for the Dol eans measure then I should write µ M 2 instead of the µ M used in Project 6. Indeed, Definition 2 would give µ M 0 if applied directly to the submartingale M. You might want to go back and make the necessary changes. <5> Example. Let {B t : 0 t 1} be a standard Brownian motion. The submartingale S t := Bt 2 has a very simple Doléans measure, characterized by µ S (a, b] F = PF ( Bb 2 a B2 = PF (b a for F Fa. That is, µ S = P m, with m equal to Lebesgue measure on B(0, 1]. Of course µ S has a further extension to the product sigma-field F B(0, 1]. A Poisson process {N t : 0 t 1} with intensity 1 shares with Brownian motion the independent increment property. It is a submartingale with sample paths that are constant, except for jumps of size 1 corresponding to points of the process; the increment N t N s has a Poisson(t s distribution. It has Doléans measure µ N = P m, the same as the square of Brownian motion. Clearly the Doléans measure does not uniquely determine the submartingale. But the only square integrable martingale M with continuous sample paths and Doléans measure µ M 2 = P m is Brownian motion: if F F s and s < t then PF (Mt 2 Ms 2 = µ M 2F (s, t] = (t spf, from which it follows that Mt 2 t is a martingale with respect to {F t }. It follows from Lévy s characterization that M is a Brownian motion.
C.2 3 C.2 Existence of the Doléans measure for class [D] There are several ways to prove that a cadlag submartingale {S t : 0 t 1} of class [D] has a countably additive Doléans measure. For example: (i Invoke an approximation by compact sets (see Métivier 1982, Chapter 3 or Chung and Williams 1990, Section 2.8, for example. (ii For the square of a continuous M 2 0 [0, 1]-martingale M, prove directly the existence of an increasing process A (the quadratic variation process for which Mt 2 A t is a martingale (Chung and Williams 1990, Section 4.4. (iii Do something very general, as in Dellacherie and Meyer (1982, 7.1. I will present a different method, based on the identification of measures on P with a certain kind of linear functional defined on the vector space H BddLip of all adapted, continuous processes H on (0, 1] Ω for which there exists a finite constant C H such that (i H(t, ω C H for all (t, ω. (ii H(s, ω H(t, ω C H t s for all s, t, and ω. The limit H(0, ω := lim t 0 H(t, ω exists for each ω. Each member of H BddLip is predictable. Also, as you saw in Project 6, H BddLip generates the sigmafield P. <6> Theorem. There exists a one-to-one correspondence between finite measures on P and increasing linear functionals µ : H BddLip R for which µ(h k 0 for each {h k : nk N} H BddLip with h k 0 pointwise. See Pollard (2001, Appendix A for a proof. Proof (of Theorem 3 Construct µ as a limit of simpler functionals. For each n in N and i = 0, 1,..., 2 n define t i,n := i/2 n. Write P i,n ( for expectations conditional on F(t i,n. Define i,n := S(t i+1,n S(t i,n. Note that P i,n i,n 0 almost surely, by the submartingale property. For each H in H BddLip, define µ n H := 0 i<2 n P ( H(t i,n i,n = 0 i<2 n P ( H(t i,n P i,n i,n. Clearly, µ n is a linear functional on H BddLip. The second expression for µ n H ensures that µ n is increasing in H.
C.2 4 The proof involves four steps: (a Show that µh := lim n µ n H exists for each H H BddLip. Note that µ inherits its linearity and the increasing property from µ n. (b For each nonnegative H H BddLip and each ɛ > 0 define τ(h, ɛ := inf{t : H(t, ω ɛ} 1. Show that µh ɛps 1 + C H P ( S 1 S τ(h,ɛ. (c Show that µ(h k 0 if h k 0 pointwise. (d Show that µ((0, τ]] = PS τ for each τ T 1. Proof of (a. For each H H, show that the sequence {µ n H : n N} is Cauchy. Fix n and m with n < m. Define J i := {j : t i,n t j,m < t i+1,n }. Then ( PH(t j,m j,m PH(t i,n i,n j Ji = j J i P ( H(t j,m H(t i,n j,m j J i P ( H(t j,m H(t i,n P j,m j,m j J i C H 2 n P j,m = C H 2 n j J i P i,n Sum over i to deduce that µ m H µ n H C H 2 n PS 1, which tends to zero as n tends to infinity. Proof of (b. Temporarily write τ n for the discretized stopping time obtained by rounding τ(h, ɛ up to the next integer multiple of 2 n. Then µ n H 0 i<2 n P ( ɛ{t i,n < τ n } + C H {t i,n τ n } P i,n i,n ɛ 0 i<2 n P i,n + C H 0 i<2 n P{t i,n τ n } i,n ɛps 1 + C H P (S 1 S τn. Let n tend to infinity. Uniform integrability of the sequence {S τn } together with right-continuity of the sample paths of S gives the asserted inequality.
Problems for Appendix C 5 Proof of (c. For a fixed ɛ > 0, temporarily write σ k for τ(h k, ɛ. By compactness of [0, 1], the pointwise convergence of the continuous functions, H k (, ω 0, is actually uniform. For each ω, the sequence {σ k (ω} not only increases to 1, it actually achieves the value 1 at some finite k (depending on ω. Uniform integrability of {S σk : k N} and the analog of (b for each H k then give µh k ɛps 1 + C H P (S 1 S σk ɛps 1 as k. By Theorem 6, the functional corresponds to the integral with respect to a finite measure on P, with total mass µ((0, 1]] = lim n µ n ((0, 1]] = PS 1. Proof of (d. For a given ɛ > 0, approximate ((0, τ]] by H ɛ (t, ω := min ( 1, (τ(ω + ɛ t + /ɛ for 0 t 1. H ε ε 0 τ τ+ε 1 Note that {(ω, t S : H ɛ (ω, t 1 c} = ((0, τ + cɛ]] for each 0 c < 1, which ensures that H ɛ is predictable. It belongs to H BddLip with C Hɛ = 1/ɛ, and µ((0, τ]] µh ɛ µ((0, τ + ɛ]]. When 2 n < ɛ, PS τ+2ɛ P i {t i,n τ + ɛ}p i,n i,n µ n H ɛ P i {t i,n τ}p i,n i,n PS τ. In the limit as n we get PS τ+2ɛ µh ɛ PS τ. As ɛ 0, the countable additivity of µ gives µ((0, τ + ɛ]] µ((0, τ]]. Property[D] and right-continuity of the sample paths gives PS τ+2ɛ PS τ. It follows that µ((0, τ]] = PS τ.
Problems for Appendix C 6 Problems [1] Let {X i : 0 i n} be a submartingale with X 0 0. For a fixed λ R +, define stopping times (i Show that (ii Show that σ := min{i : X i λ} 1 and τ := min{i : X i λ} 1. λp{max i X i > λ} PX τ {X τ λ} PX 1 {X τ λ} P X n. λp{min i X i < λ} P( X σ {X σ λ} PX σ + PX n {X σ > λ} P X n. (iii Suppose {Y t : 0 t 1} is a cadlag submartingale with Y 0 0. Show that λp{sup t Y t > λ} 2P Y 1. [2] Suppose a cadlag martingale {S t : 0 t 1}, with S 0 0, has a Doléans measure µ in the sense of Definition <2>, that is, µ((0, τ]] = PS τ for every τ T 1. Show that S has property [D] by following these steps. (i For a given τ T 1, let τ n be the stopping time obtained by rounding up to the next integer multiple of 2 n. (ii Invoke the Stopping Time Lemma to show that 0 PS τn and PS + τ n PS + 1 for each τ T 1. Deduce that P S τn κ := 2PS + 1 <. (iii Invoke Fatou s lemma to show that sup τ T1 P S τ κ. (iv For each C R +, show that PS τn {S τn > C} PS 1 {S τn > C} PS 1 {S 1 > C} + κ/ C. Invoke Fatou, then deduce that sup τ T1 PS τ {S τ > C} 0 as C. (v Show that every cadlag function on [0, 1] is bounded in absolute value. Deduce that the stopping time σ C := inf{t : S t < C} 1 has σ C (ω = 1 for all C large enough (depending on ω. Deduce that µ((σ C, 1]] 0 as C. (vi For a given τ T 1 and C R +, define F τ := {S τ < C}. Show that τ := τfτ c + F τ is a stopping time for which P ( S 1 S τ Fτ = P ( S τ S τ = µ((τ, τ ]] µ((σ C, 1]],
References for Appendix C 7 Hint: Show that if ω F τ then σ C (ω τ(ω and if ω F c τ then τ(ω = τ (ω. (vii Deduce that sup τ T1 P ( S τ {Sτ < C} 0 as C. [3] Let {S t : t R + } be a submartingale of class [D]. Show that there exists an integrable random variable S for which P ( S F t St S almost surely and in L 1 by following these steps. (i Show that the uniformly integrable submartingale {S n : n N} converges almost surely and in L 1 to an S for which P ( S F n Sn. (ii For t n, show that S t P ( P(S F n F t = P ( S F t. (iii For t n, show that P(S t S n P(S t S n + P(S S n + 0 as n. (iv For each k N, choose n(k for which P S S n(k 4 k. Invoke Problem [1] to show that k P{sup t n(k S t S n(k > 2 k } <. (v Deduce that S t S almost surely. C.3 Notes My exposition in this Chapter is based on ideas drawn from a study of Métivier (1982, 13, Dellacherie and Meyer (1982, Chapter VII, and Chung and Williams (1990, Chapter 2. The construction in Section 2 appears new, although it is clearly closely related to existing methods. References Chung, K. L. and R. J. Williams (1990. Introduction to Stochastic Integration. Boston: Birkhäuser. Dellacherie, C. and P. A. Meyer (1982. Probabilities and Potential B: Theory of Martingales. Amsterdam: North-Holland. Métivier, M. (1982. Semimartingales: A Course on Stochastic Processes. Berlin: De Gruyter. Pollard, D. (2001. A User s Guide to Measure Theoretic Probability. Cambridge University Press.